solutions mht cet-2018 paper-i (code-22) paper-i final cet-2018... · = sec log cosec cotx xxc 4....

1

Answers & Solutionsfor for for for for MHT CET-2018

Paper-I

(Mathematics)

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005

Ph.: 011-47623456 Fax : 011-47623472

DATE : 10/05/2018

Instruction for Candidates

1. This question booklet contains 50 Objective Type Questions (Single Best Response Type) in the subject of

Mathematics (50).

2. The question paper and OMR (Optical Mark Reader) Answer Sheet are issued to examinees separately at

the beginning of the examination session.

3. Choice and sequence for attempting questions will be as per the convenience of the candidate.

4. Read each question carefully.

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6. Each answer with correct response shall be awarded two (2) marks. There is no Negative Marking. If the

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sheet of each candidate will be evaluated by computerized scanning method only (Optical Mark Reader)

and there will not be any manual checking during evaluation or verification.

7. Rough work should be done only on the blank space provided in the Question Booklet. Rough work should

not be done on the Answer Sheet.

8. The required mathematical tables (Log etc.) are provided within the Question Booklet.

Time : 1 Hour 30 Min. Total Marks : 100

Question Booklet Version

22

2

MHT-CET - 2018 (Paper-I) Code-22

1. If 2

0242 18

Kdx

x

∫ , then the value of K is

(A) 3 (B) 4

(C)1

3(D)

1

4

Answer (C)

Sol. 2

0

2

0

1

22 18 (

d d

) 13

KK

x

x x

x

⌠⌠⎮ ⎮⌡ ⌡

1

0

1tan 3

6

K

x ⎤ ⎥

⎦

11tan 3

6 24K

hence 1 1tan

3 4 3K

2. The cartesian co-ordinates of the point on the

parabola y2 = –16x, whose parameter is 1

2, are

(A) (–2, 4) (B) (4, –1)

(C) (–1, –4) (D) (–1, 4)

Answer (C, D)*

Sol. *For the curve y2 = –16x, the parametric form of

coordinates of any point are (–4t2, 8t). Hence for

1

2t , point is (–1, 4).

3. dxx x

2

1

sin .cos∫

(A) x x x csec log sec tan

(B) x x csec .tan

(C) x x x csec log sec tan

(D) sec log cosec cotx x x c

Answer (D)

Sol. The given integral can be written as

2 2

2

sin cosd

sin cos

x xx

x x

⌠⎮⌡

= sec tan cosecx x x dx∫

= sec log cosec cotx x x c

4. If x y

x y

3 3

10 3 3log 2

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠then

dy

dx =

(A)x

y(B)

y

x

(C)x

y (D)

y

x

Answer (D)

Sol. Taking antilog,

3 32

3 310

yx

x y

.

Applying componendo and dividendo,

3 3 3 3

3 3 3 3

) ( ) 100 1

100 1) ( )

(

(

x

x

y x y

y x y

Simplifying, we have

33 99

101

xy

.

Differentiating,

2

2 993

1

3

01

dy xy

dx

2

2

99

101

dy x

dx y

3 2

3 2

dy y x y

dx xx y

5. If f : R – {2} R is a function defined by

xf x

x

2 4( )

2

, then its range is

(A) R (B) R – {2}

(C) R – {4} (D) R – {–2, 2}

Answer (C)

Sol. The function can be simplified as 2, 2y x x .

The value corresponding to x = 2 is not in the

range. Hence Range is R – {4}

6. If f x x2( ) for 0x

= x2

2 1 for x 0

is continuous at x = 0 and f1

22

⎛ ⎞ ⎜ ⎟⎝ ⎠

then

2 2 is

(A) 3 (B)8

25

(C)25

8(D)

1

3

MATHEMATICS

3


Answer (C)

Sol. Since the function f(x) is continuous, we must have

0 0

lim lim( ) ( ) (0)x x

f x f x f

But 2

0 0

( ) 2lim lim 1 2 , (0)x x

f x x f

and 2

0 0

lim li( m)x x

f x x

Hence 2 + = . ....(i)

1 1 7 12 ,

2 4 4 4f⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

2 2 25

8⇒

7. If y x2

–1tan then

d y dyx x x

dxdx

22

2 2

21 2 1

(A) 4 (B) 2

(C) 1 (D) 0

Answer (B)

Sol. Differentiating the given function we get1

2

tan2

1

dy x

d xx

. Differentiating second time, we get

2 1

2 22

ta2 4

1

nx

x

d y x

dx

. Rearranging the last

equation we have, 2

22 1

21 2 4 tan

d yxx x

dx

.

Substituting 2

1 (1t

2

)an

x dyx

dx

, we finally get

2

2 2 2

2(1 ) 2 (1 ) 2

d y dyx x x

dxdx

8. The line 5x + y – 1 = 0 coincides with one of the

lines given by 5x2 + xy – kx – 2y + 2 = 0 then the

value of k is

(A) –11 (B) 31

(C) 11 (D) –31

Answer (C)

Sol. If the lines represented by pair of lines ax2 + by2 +

2gx + 2fy + 2hxy + c = 0 has slopes m1 and m

2,

then we have m1 + m

2 =

2h

b , which in this case

is not defined as b = 0.Since slope of one of the

lines is given as m = –5, the slope of other must

be infinite. Hence the other line must be of the form

x = k1. Comparing term by term (5x + y – 1)(x –

k1) 5x2 + xy – kx – 2y + 2, we get k = 11

9. If A

1 2 3

–1 1 2

1 2 4

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

then A A A2 –1

5 =

(A)

4 2 3

–1 4 2

1 2 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(B)

–4 2 3

–1 –4 2

1 2 –1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(C)

–4 –1 1

2 –4 2

3 2 –1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(D)

–1 –2 1

4 –2 –3

1 4 –2

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Answer (B)

Sol. 2 155A A A A I

=

1 2 3 5 0 0

1 1 2 0 5 0

1 2 4 0 0 5

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

4 2 3

1 4 2

1 2 1

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

10. The equation of line passing through (3, –1, 2) and

perpendicular to the lines

r i j k i j kˆ ˆ ˆ ˆ ˆ ˆ– 2 – 2 and

r i j k i j kˆ ˆ ˆ ˆ ˆ ˆ2 – 3 – 2 2 is

(A)x y z3 1 2

2 3 2

(B)x y z3 1 2

3 2 2

(C)x y z3 1 2

2 3 2

(D)x y z3 1 2

2 2 3

Answer (C)

Sol. The direction vector of the line is given by

2 2 1 2

1 2 2

ˆ ˆ ˆ

ˆ ˆ ˆ3 2

i j k

i j k

. Since line passes

through (3, –1, 2), its equation is

3 1 2

2 3 2

x y z

4


11. Letters in the word HULULULU are rearranged. The

probability of all three L being together is

(A)3

20(B)

2

5

(C)3

28(D)

5

23

Answer (C)

Sol. size of the sample space, n(S) = 8!

4! 3!

number of favorable cases, n(E) = 6!

4!

P = n

n

( )

(S)

E 6! 4! 3!

4! 8! =

6 3

8 7 28

12. The sum of the first 10 terms of the series

9 99 999 ..... is

(A) 1099 1

8 (B) 9100

10 19

(C) 910 1 (D) 10100

10 19

Answer (B)

Sol. 9 + 99 + 999 +

[(10 – 1) + (102 –1 ) + (103 – 1) + 10th term]

[(101 + 102 + 103 + 10th term) – 10]

=

1010 1–10–10

1–10

⎡ ⎤⎡ ⎤⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

=

10

1010 –1 10

10 –10 10 –1 – 99 9

⎡ ⎤⎡ ⎤⎢ ⎥ ⎣ ⎦

⎢ ⎥⎣ ⎦

= 9100

10 –19

⎡ ⎤⎣ ⎦

13. If A, B, C are the angles of ABC then

cotA.cotB + cotB.cotC + cotC.cotA =

(A) 0 (B) 1

(C) 2 (D) –1

Answer (B)

Sol. Since tan(A + B + C) = tan = 0.

tan tan tan – tan tan tan0

1 tan tan – tan tan – tan tan

A B C A B C

A B B C C A

Thus

tan tan tan – tan tan tan 0A B C A B C

Dividing by tan A tan B tan C and simplifying we

get

cot A cot B + cot B cot C + cot C cot A =1

14. If –1

2

sin

16 9

dxA Bx C

x

∫ then A + B =

(A)9

4(B)

19

4

(C)3

4(D)

13

12

Answer (D)

Sol.2

2

1

3 1616 – 9–

9

dx dx

xx

∫ ∫ –11 3sin

3 4

xC

⎛ ⎞ ⎜ ⎟⎝ ⎠

Comparing, we get A = 1

3, B =

3

4

Therefore A + B = 1 3 4 9 13

3 4 12 12

15.x

xe dx

x

2 sin2

1 cos2

⎡ ⎤ ⎢ ⎥⎣ ⎦∫

(A)x

e x ctan (B)x

e x ctan

(C)x

e x c2 tan (D)x

e x ctan2

Answer (A)

Sol.2 sin(2 )

1 cos2x

xe dx

x

⎡ ⎤⎢ ⎥⎣ ⎦

∫

=2

2 2sin cos

2cos

xx x

e dxx

⎡ ⎤⎢ ⎥⎣ ⎦

∫

= 2 tansecx xe x dx⎡ ⎤⎣ ⎦∫= extanx + c

16. A coin is tossed three times. If X denotes the

absolute difference between the number of heads

and the number of tails then P(X = 1) =

(A)1

2(B)

2

3

(C)1

6(D)

3

4

5


Answer (D)

Sol. There are two possible favourable cases: HHT or

TTH. Where H denotes Head and T denotes Tail.

P(X = 1) =

3

31

12

2C

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 3

4

17. If 2sin cos3 6

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

then tan=

(A) 3 (B)1

3

(C)1

3(D) 3

Answer (D)

Sol. 2sin cos –3 6

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

sin 3 cos 3 cos sin2

2 2 2 2

⎡ ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦

2sin 2 3 cos 3 cos sin

sin – 3 cos

tan = – 3

18. The area of the region bounded by x2 = 4y, y = 1,

y = 4 and the y-axis lying in the first quadrant is

________ square units.

(A)22

3(B)

28

3

(C) 30 (D)21

4

Answer (B)

Sol. y = 2

4

x

y = 4

y = 1

Required area =

4

1

2 ydy∫ =

4

3/2

13

22

3

28y

⎤ ⎥⎦

19. If x

e xf x

x

2

2

cos( )

, for x 0 is continuous at

x = 0, then value of f(0) is

(A)2

3(B)

5

2

(C) 1 (D)3

2

Answer (D)

Sol. Apply L'Hospital Rule,

2

20

– coslim

x

x

e x

x=

2

0

2 sinlim

2

x

x

xe x

x

2

0

sin 1 3lim 1

2 2 2

x

x

xe

x

20. The maximum value of 2x + y subject to

3x + 5y 26 and 5x + 3y 30, x 0, y 0 is

(A) 12 (B) 11.5

(C) 10 (D) 17.33

Answer (A)

Sol.

(0, 10)

260,

5

⎛ ⎞⎜ ⎟⎝ ⎠

9 5,

2 2

⎛ ⎞⎜ ⎟⎝ ⎠

(6, 0)26

,03

⎛ ⎞⎜ ⎟⎝ ⎠

3 + 5 = 26x y

5 + 3 = 30x y

Evaluating the function 2x + y at the corner points

of the shaded region, it can easily be seen that the

maximum occurs at the point (6,0) and is 12.

21. If , , a b c�

� �

are mutually perpendicular vectors having

magnitudes 1, 2, 3 respectively, then

a b c b a c⎡ ⎤ ⎣ ⎦

� ��

=

(A) 0 (B) 6

(C) 12 (D) 18

Answer (C)

Sol. a b c b a c⎡ ⎤ ⎣ ⎦

� ��

= ( ) (( ) )a b c b a c � ��

i

= ( ) ( )a b c ccb a � � ��

i

= [ ] [ ]ab c b ca ��

��

= 2[ ]a b c�

� �

= 2 1 2 3 12

6


22. If points P(4, 5, x), Q(3, y, 4) and R (5, 8, 0) are

collinear, then the value of x + y is

(A) –4 (B) 3

(C) 5 (D) 4

Answer (D)

Sol.4 3 5 4

3 5 8 4

y x

y

1 5

2 8

y

y

8 10 2y y y = 2

1 4

2 4

x

– 2 = x – 4 x = 2

x + y = 2 + 2 = 4

23. If the slope of one of the lines given by ax2 + 2hxy

+ by2 = 0 is two times the other then

(A) 8h2 = 9ab

(B) 8h2 = 9ab2

(C) 8h = 9ab

(D) 8h = 9ab2

Answer (A)

Sol. Let the slopes of lines be m, 2m

sum of roots 2

2h

m mb

23

hm

b

⇒ ...(i)

product of roots 22

am

b ...(ii)

eliminating m from both equations we get

28 9h ab

24. The equation of the line passing through the point

(–3, 1) and bisecting the angle between co-ordinate

axes is

(A) x + y + 2 = 0

(B) –x + y + 2 = 0

(C) x – y + 4 = 0

(D) 2x + y + 5 = 0

Answer **

Sol. **Wrong Question

The lines bisecting the angle between coordinate

axes are y = x and y = –x which do not passes

through the point (–3, 1)

25. The negation of the statement : “Getting above 95%

marks is necessary condition for Hema to get the

admission in good college”.

(A) Hema gets above 95% marks but she does not

get the admission in good college

(B) Hema does not get above 95% marks and she

gets admission in good college

(C) If Hema does not get above 95% marks then

she will not get the admission in good college

(D) Hema does not get above 95% marks or she

gets the admission in good college

Answer (B)

Sol. Let A denote "Hema to get the admission in good

college"

B denote "Getting above 95% marks"

Then given statement is A B. Negation of this

statement is

) ~ (~ A ) ~~ (A B B A B

Hence the required negation in words is "Hema

does not get above 95% marks and she gets

admission in good college”.

26. cos1°cos2°cos3°cos179° =

(A) 0 (B) 1

(C)1

2 (D) –1

Answer (A)

Sol. cos 90° = 0

27. If planes x – cy – bz = 0, cx – y + az = 0 and

bx + ay – z = 0 pass through a straight line then

a2 + b2 + c2 =

(A) 1 – abc (B) abc – 1

(C) 1 – 2abc (D) 2abc – 1

Answer (C)

Sol.

1

1 0

1

c b

c a

b a

21(1 ) ( ) ( ) 0a c c ab b ac b

2 2 21 0a c abc abc b

2 2 21 2a b c abc

28. The point of intersection of lines represented by

x2 – y2 + x + 3y – 2 = 0 is :

(A) (1, 0) (B) (0, 2)

(C)1 3,

2 2

⎛ ⎞⎜ ⎟⎝ ⎠

(D)1 1,

2 2

⎛ ⎞⎜ ⎟⎝ ⎠

7


Answer (C)

Sol. Let s = x2 – y2 + x + 3y – 2

12 1 0

2

dsx x

dx ⇒

32 3 0

2

dsy y

dy ⇒

29. A die is rolled. If X denotes the number of positive

divisors of the outcome then the range of the

random variable X is :

(A) {1, 2, 3}

(B) {1, 2, 3, 4}

(C) {1, 2, 3, 4, 5, 6}

(D) {1, 3, 5}

Answer (B)

Sol.

1 {1} 1

2 {1,2} 2

3

Outcomes Divis

{1,3} 2

4 {1,2,4

ors Number

} 3

5 {1,5} 2

of Divisors

6 {1,2,3,6} 4

Hence range of X is {1, 2, 3, 4}.

30. A die is thrown four times. The probability of getting

perfect square in at least one throw is :

(A)16

81(B)

65

81

(C)23

81(D)

58

81

Answer (B)

Sol. Required Probability = 1 – P(No perfect square in

any throw)

Perfect square are 1, 4. Probability of perfect

square in any one throw = 2

6 3

1

P(at least one perfect. square)

42

13

⎛ ⎞ ⎜ ⎟⎝ ⎠

16 81 161

81 81

65

81

31.4

2

0

.sec .x x dx

∫

(A) log 24

(B) log 24

(C) 1 log 2 (D)1

1 log22

Answer (B)

Sol. Using integration by parts

4

2

0

secx x dx

∫ = 4

0

0

4 – ttan anx xdxx

∫

= – log 24

⎡ ⎤⎢ ⎥⎣ ⎦

32. In ABC, with usual notations, if a, b, c are in A.P.

then a cos2

2

C⎛ ⎞⎜ ⎟⎝ ⎠

+ c cos2

2

A⎛ ⎞⎜ ⎟⎝ ⎠

=

(A) 32

a

(B) 32

c

(C) 32

b(D)

3

2

abc

Answer (C)

Sol. co1 o2 2

s 1 c sCa c

A

= cos cos

2 2 2

a c a C c A

3

2 2

a b c b

33. If x = e(sin – cos ), y = e(sin + cos ) then

dy

dx at =

4

is

(A) 1 (B) 0

(C)1

2(D) 2

Answer (A)

Sol. x = e (sin – cos)

y = e (sin + cos)

cos – sin sin cos

cos sin sin – cos

e edy

dx e e

4

1dy

dx

8


34. The number of solutions of sin x + sin 3x + sin 5x

= 0 in the interval 3

, 2 2

⎡ ⎤⎢ ⎥⎣ ⎦

is

(A) 2 (B) 3

(C) 4 (D) 5

Answer (B)

Sol. sinx + sin3x + sin5x = 0

2sin 3x cos2x + sin3x = 0

sin3x [2cos2x + 1] = 0

either sin3x = 0 or 2cos 2x + 1 = 0

case (1), sin 3x = 0, then 3x = n

x = 3

n, n is any integer

If solution lies in the interval 3,

2 2

⎡ ⎤⎢ ⎥⎣ ⎦

, then n = 2,

n = 3, n = 4, 2 4

, , 3 3

x ⎧ ⎫ ⎨ ⎬

⎩ ⎭

Case (2), cos2x = 1 2

– cos2 3

Then 2x = 2

2 ,3

n n �

x = 3

n , which gives two solutions

2 4,

3 3

Hence total 3 solutions exist in the interval.

35. If tan–1 2x + tan–1 3x = 4

, then x =

(A) –1 (B)1

3

(C)1

6(D)

1

2

Answer (C)

Sol. –1 –1tan 2 tan 34

x x

–1

2

5tan

1– 6 4

x

x

2

51

1– 6

x

x

6x2 + 5x – 1 = 0

6x2 + 6x – x – 1 = 0

6x(x + 1) – (x + 1) = 0

(6x – 1) (x + 1) = 0

x = 1, 1

6x

But x = –1 does not satisfy the given equation.

x = 1

6

36. Matrix A =

1 2 3

1 1 5

2 4 7

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

then the value of a31

A31

+

a32

A32

+ a33

A33

is

(A) 1 (B) 13

(C) –1 (D) –13

Answer (C)

Sol. It is ambiguous that is denoted by aij.

Assuming aij to be elements of co-factor matrix we

have a31

= (–1)3 + 1 (52 – 13) = 7, a32

= (–1)3 + 2

(51 – 13) = –2, a33

= (–1)3 + 3 (11 – 12) = –1.

Hence a31

A31

+ a32

A32

+ a33

A33

= 72 + (–24) +

(–17) = –1.

37. The contrapositive of the statement : “If the weather

is fine then my friends will come and we go for a

picnic.”

(A) The weather is fine but my friends will not come

or we do not go for a picnic

(B) If my friends do not come or we do not go for

picnic then weather will not be fine

(C) If the weather is not fine then my friends will

not come or we do not go for a picnic

(D) The weather is not fine but my friends will come

and we go for a picnic

Answer (B)

Sol. Let A denote "Weather is fine"

B denote "my friends will come"

C denote "we go for a picnic"

Then given statement is A B C . To find its

contrapositive, we need to evaluate

)~ ~ A(B C i.e., )~ ~ ~( C AB .

In words this becomes "If my friends do not come

or we do not go for picnic then weather will not be

fine".

9


38. If 2

( )1

xf x

x

is increasing function then the value

of x lies in

(A) R (B) ( , 1)

(C) (1, ) (D) (–1, 1)

Answer (D)

Sol. f(x) = 2 1

x

x

2

22

( 1)1( ) 0

1

(2 )x xf x

x

x

1 – x2 > 0

x2 – 1 < 0

x (–1, 1)

39. If (4 3 1: )n

X n n N and {9( 1) : )Y n n N ,

then X Y ∩

(A) X (B) Y

(C) (D) {0}

Answer (A)

Sol. Consider set X, its element are of the form

4n – 3n – 1

= (1 + 3)n – 3n – 1

= 21 21 3 3 – 3 – 1n nC C n

= 2 –223 3n n n

nC C⎡ ⎤ ⎣ ⎦ 9p

Set Y consists of all non negative multiples of 9.

Hence every element of set X is a subset of Y.

40. The statement pattern (~ )p p q is

(A) A tautology

(B) A contradiction

(C) Equivalent to p q

(D) Equivalent to p q

Answer (B)

Sol. Consider the truth table of ~p p q

~ )~ (~p q p p p

T T F F F

T F F F F

F T T T F

q p

T

q

F F F F

Hence a Contradiction.

41. If the line y = 4x – 5 touches to the curve y2 = ax3

+ b at the point (2, 3) then 7a + 2b =

(A) 0 (B) 1

(C) –1 (D) 2

Answer (A)

Sol. y2 = ax3 + b ...(1)

Since (2,3) lies on the curve, we have

9 = 8a + b ...(2)

Differentiating (1),

2

2 32 3

2 ⇒ dy dy ax

y axdx dx y

Since line y = 4x – 5 touches the curve

3 44 2 –7

2 3

⇒ ⇒ ⇒

a

a b

7a + 2b = 0

42. The sides of a rectangle are given by x = ±a and

y = ±b. The equation of the circle passing through

the vertices of the rectangle is

(A) x2 + y2 = a2

(B) x2 + y2 = a2 + b2

(C) x2 + y2 = a2 – b2

(D) (x – a)2 + (y – b)2 = a2 + b2

Answer (B)

Sol. (a, b) and (–a, –b) will be ends of diameter.

Required equation of circle is

(x + a) (x – a) + (y + b) (y – b) = 0

x2 + y2 = a2 + b2

43. The minimum value of the function f(x) = xlogx is

(A)1

–e

(B) –e

(C)1

e(D) e

Answer (A)

Sol. f(x) = xlogx

1'( ) 1 log '( ) 0 ⇒ ⇒

ef x x f x x

e

and 1 1

"( ) " 0⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

f x fx e

minima exists.

Therefore minimum value of f(x) will occur at 1x

e

1 1 1 1log –

⎛ ⎞⇒ ⎜ ⎟

⎝ ⎠f

e e e e

10


44. If X ~ B (n, p) with n = 10, p = 0.4 then E(X2) =

(A) 4 (B) 2.4

(C) 3.6 (D) 18.4

Answer (D)

Sol. Var(X) = E(X2) – (E(X))2

E(X2) = Var(X) + (E(X))2

= npq + (np)2

= (10 × .4 × .6) + (10 × .4)2

= 2.4 + 16

= 18.4

45. The general solution of differential equation

cos( ) dxx y

dy is

(A) tan2

x yy c

⎛ ⎞ ⎜ ⎟⎝ ⎠

(B) tan2

x yx c

⎛ ⎞ ⎜ ⎟⎝ ⎠

(C) cot2

x yy c

⎛ ⎞ ⎜ ⎟⎝ ⎠

(D) cot2

x yx c

⎛ ⎞ ⎜ ⎟⎝ ⎠

Answer (A)

Sol. cos( ) dxx y

dy

Let x + y = t

1dx dt

dy dy⇒

–1 cosdt

tdy

1 cos⇒ dtt

dy

22cos

2

dt t

dy⇒

2

2

cos2

dtdy

t⇒

2sec 2

2

tdt dy⇒ ∫ ∫

2tan 22

⇒ ty c

2tan 22

⇒ x y

y c

tan2

⇒ x y

y c

46. If planes � . – 2 3 0r pi j k

� � � and

� . 2 – – – 5 0r i p j k � � �

include angle 3

then the

value of p is

(A) 1, –3

(B) –1, 3

(C) –3

(D) 3

Answer (D)

Sol.2 2

2 – 2cos60

5 5

p p

p p

(3p – 2)2 = p2 + 5

p2 – 6p + 9 = 0

(p – 3)2 = 0

p = 3

47. The order of the differential equation of all parabolas,

whose latus recturn is 4a and axis parallel to the

x-axis, is

(A) one

(B) four

(C) three

(D) two

Answer (D)

Sol. Required equation of parabola is

(y – m)2 = 4a (x – n)

Here m and n are two independent arbitrary

constant.

Order of required differential equation is 2.

48. If lines –1 1 –1

2 3 4

x y z and –

– 32

y kx z

intersect then the value of k is

(A)9

2(B)

1

2

(C)5

2(D)

7

2

11


Answer (A)

Sol. General point on these lines are

(2 + 1, 3 – 1, 4 + 1) and ( + 3, 2 + k, )

For point of intersection

2 + 1 = + 3 and 4 + 1 =

3

2 and = – 5

and 9

3 –1 2 – –1 –102

k k ⇒

11 910 –

2 2k k⇒ ⇒

49. If a line makes angles 120° and 60° with the

positive directions of X and Z axes respectively then

the angle made by the line with positive Y-axis is

(A) 150° (B) 60°

(C) 135° (D) 120°

Answer (C)

Sol. Let the angles which the line makes with positive

directions of X, Y and Z axes be , and Then

= 120°

= 60°

= ?

cos2 + cos2 + cos2 = 1

21 1cos 1

4 4⇒

2 1 1 1cos cos or –

2 2 2⇒ ⇒

Hence = 45° or 135°

50. L and M are two points with position vectors 2 –a b�

�

and 2a b�

�

respectively. The position vector of the

point N which divides the line segment LM in the

ratio 2 : 1 externally is

(A) 3b

�

(B) 4b

�

(C) 5b

�

(D) 3 4a b�

�

Answer (C)

Sol. Let the position vector of point N be n�

. Then

n�

= 2( 2 ) 1(2 )

2 1

a b a b

� ��

= 5b�

� � �

solutions mht cet-2018 paper-i (code-22) paper-i final cet-2018... · = sec log cosec cotx xxc 4....

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